Answer:

We are given that –

Angle of minimum deviation is  $\delta_{m}=40°$

Angle of the prism is  A = $60°$

Refractive Index of water is given by, $\mu=1.33$

Material’s refractive Index = $\mu ‘$

We know that the angle of deviation and the refractive index ($\mu ‘$) are related to each other as –

$\mu ‘$=$\frac{sin\frac{A+\delta_{m}}{2}}{sin\frac{A}{2}}$

=$\frac{sin\frac{60°+40°}{2}}{sin\frac{60°}{2}}=\frac{si n 50°}{sin 30°}$

= 1.532

1.532 is the refractive index of the prism material.

Since the prism has been immersed in water, $\delta_{m}$ is the new angle of minimal deviation. The graph depicts the refractive index of glass in relation to water.

$\delta_{g}^{w}=\frac{\mu ‘}{\mu}=\frac{sin\frac{(A+\delta_{m}’)}{2}}{sin\frac{A}{2}} $

$ sin\frac{(A+\delta_{m}’)}{2}=\frac{\mu ‘}{\mu}sin\frac{A}{2}$

$ sin\frac{(A+\delta_{m}’)}{2}=\frac{1.532}{1.33}\times sin\frac{60°}{2}=0.5759$

$ \frac{(A+\delta_{m}’)}{2}=sin^{-1}0.5759=35.16°$$ 60°+\delta_{m}’=70.32°$

$ Therefore\;\delta_{m}’=70.32°-60°=10.32°$

$10.32°$ is the calculated new minimum angle of deviation.